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Find the Frequency of Revolution of an Electron in Bohr’S 2nd Orbit; If the Radius and Speed of Electron in that Orbit is 2.14 x 10^-10 M and 1.09 x 10^6 M/S Respectively. - HSC Science (General) 12th Board Exam - Physics

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Question

Find the frequency of revolution of an electron in Bohr’s 2nd orbit; if the radius and speed of electron in that orbit is 2.14 × 10-10 m and 1.09 × 106 m/s respectively. [π= 3.142]

Solution 1

Given

r2 = 2.14 × 10-10 m

n = 2

v2 = 1.09 × 106 m/s

To find: Frequency of revolution (`nu_2`)

`v = romega=r(2pinu)`

`nu=v/(2pir)`

`nu_2=v_2/(2pir_2)=(1.09xx10^6)/(2xx3.142xx2.14xx10^-10)`

`nu_2=8.11xx10^14 Hz`

The frequency of revolution of electron in 2nd Bohr orbit is 8.11 × 1014 Hz.

Solution 2

`T = (2pir)/V`

`because T=1/f`

`f = V/(2pir)`

`f=(1.09 xx 10^6)/(2 xx 3.14 xx 2.14 xx 10^-10)`

`f=8.11 xx 10^14 Hz`

The frequency of revolution of electron in 2nd Bohr orbit is 8.11 × 1014 Hz.

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APPEARS IN

 2016-2017 (March) (with solutions)
Question 6.8 | 2.00 marks
Solution Find the Frequency of Revolution of an Electron in Bohr’S 2nd Orbit; If the Radius and Speed of Electron in that Orbit is 2.14 x 10^-10 M and 1.09 x 10^6 M/S Respectively. Concept: Bohr'S Model for Hydrogen Atom.
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